In a combustion of a hydrocarbon compound, 2 reactions are happening per element:
C + O₂ → CO₂
2 H + 1/2 O₂ → H₂O
Thus, we can determine the amount of C and H from the masses of CO₂ and H₂O produced, respectively.
1.) Compute for the amount of C in the compound. The data you need to know are the following:
Molar mass of C = 12 g/mol
Molar mass of CO₂ = 44 g/mol
Solution:
0.5008 g CO₂*(1 mol CO₂/ 44 g)*(1 mol C/1 mol CO₂) = 0.01138 mol C
0.01138 mol C*(12 g/mol) = 0.13658 g C
Compute for the amount of H in the compound. The data you need to know are the following:
Molar mass of H = 1 g/mol
Molar mass of H₂O = 18 g/mol
Solution:
0.1282 g H₂O*(1 mol H₂O/ 18 g)*(2 mol H/1 mol H₂O) = 0.014244 mol H
0.014244 mol H*(1 g/mol) = 0.014244 g H
The percent composition of pure hydrocarbon would be:
Percent composition = (Mass of C + Mass of H)/(Mass of sample) * 100
Percent composition = (0.13658 g + 0.014244 g)/(<span>0.1510 g) * 100
</span>Percent composition = 99.88%
2. The empirical formula is determined by finding the ratio of the elements. From #1, the amounts of moles is:
Amount of C = 0.01138 mol
Amount of H = 0.014244 mol
Divide the least number between the two to each of their individual amounts:
C = 0.01138/0.01138 = 1
H = 0.014244/0.01138 = 1.25
The ratio should be a whole number. So, you multiple 4 to each of the ratios:
C = 1*4 = 4
H = 1.25*4 = 5
Thus, the empirical formula of the hydrocarbon is C₄H₅.
3. The molar mass of the empirical formula is
Molar mass = 4(12 g/mol) + 5(1 g/mol) = 53 g/mol
Divide this from the given molecular weight of 106 g/mol
106 g/mol / 53 g/mol = 2
Thus, you need to multiply 2 to the subscripts of the empirical formula.
Molecular Formula = C₈H₁₀
For reaction
2 A + B ------------> 2 C
Rate = K [ A ]² [ B ]
<span> the order with respect to A is 2 and the order overall is 3.
</span>
hope this helps!
Answer:
2KClO3 》》2KCl +3O2
C+ O2》》CO2
number of C moles
Required O2 moles (According to the mole ratio )
Relevant to the first equation, find the moles the KClO3, which is used to produce that amount of O2 moles
Now you can find the mass of KClO3
I mentioned the useful steps which can guide you to get the answer.
Explanation:
Answer:
72.22 g
Explanation:
975 mL Mercury× 13.5 g/mL = 72.22 g
Explanation:
As it is known that molarity is the number of moles present in a liter of solution.
Mathematically, Molarity = 
As it is given that molarity is 0.10 M and volume is 10.0 ml. As 1 ml equals 0.001 L. Therefore, 10.0 ml will also be equal to 0.01 L.
Hence, putting these values into the above formula as follows.
Molarity =
0.10 M = 
no. of moles = 0.001 mol
As molar mass of KCN is equal to 65.12 g/mol. Therefore, calculate the mass of KCN as follows.
No. of moles = 
0.001 mol = 
mass = 0.06152 g
Thus, we can conclude that 0.06152 grams of KCN are in 10.0 ml of a 0.10 M solution.
